A poster is to have an area of 210 in2 with 1 inch margins at the bottom and sides and a 2 inch margin at the top. Find the exact dimensions that will give the largest printed area.

Question

width=_____

height=_____

Answer 1

1) You need an expression for the printed area that you want to maximize given the h and w:

A = (h-3)(w-2) 2+1 inch margins on from height dimension, and 2 1 inch margins from width.

2) You need a constraint that allows you to eliminate one of your variables:

A of poster = hw, so h = 210/w

3) Make printed area A(w only) = (210/w - 3)(w-2)

4) Find critical points: dA/dw = 0 in order to find maximum. There will be a max as 2 is an endpoint for w and 3 is an endpoint for h (w max is 70) and both of these have printable areas that are 0,

I get w = 36. You want to solve for critical points of f(w) in the form of a quadratic: w² - 72w +140 = f(w) =

-wA(w)

5) Check if it's a max using f''(w) = 2 . Oh no it's a min. That's ok, because f(x) has the opposite sign, so it's min is A's max because w is >0 and the negative sign means that f always goes in the opposite direction as A. If you hate dealing with that, then you could have taken the messier derivative of A(w) = 216 -3w-420/w and set that equal to 0. The f(x) form is fine for finding zeroes and remember that there must be a maximum in the interval from the end points and Rolle's Theorem.

Take care.

Answer 2

wh = 210

The printable area P = (h-3)(w-2) (all borders 1 inch but 1 that is 2 on the height).

We can either substitute right off or use lagrange multipliers.

Let's just substitute h = 210/w

Then, P = (210/w - 3)(w - 2) = 210 - 420/w - 3w + 6 = 216 - 420/w - 3w

dP/dw = 420/w^2 - 3

420/w^2 = 3

3w^2 = 420

w^2 = 140

w = sqrt(140) = 2 sqrt(35)

h = 210/w = 210/(2 sqrt(35)) = 3 sqrt(35)

P'' = -840/w^3 is negative for all positive w assures a maximum where the first derivative equals 0.